Find the least value of ‘a’ so that $f(x) = 2x^2 - ax + 3$ is an increasing function on $[2, 4]$.
Show Hint
To find the value of $a$ so that a function is increasing, check the derivative and ensure it is positive over the entire interval. For quadratic functions, focus on the lowest value in the interval.
We are given the function:
\[
f(x) = 2x^2 - ax + 3.
\]
To determine the values of $a$ for which the function is increasing on $[2, 4]$, we need to find the derivative of $f(x)$:
\[
f'(x) = \frac{d}{dx} \left( 2x^2 - ax + 3 \right) = 4x - a.
\]
For the function to be increasing on $[2, 4]$, the derivative must be positive for all $x$ in this interval, i.e.,
\[
f'(x)>0 \quad \text{for all } x \in [2, 4].
\]
Thus,
\[
4x - a>0 \quad \text{for all } x \in [2, 4].
\]
At the minimum value of $x = 2$, we have:
\[
4(2) - a>0 \quad \Rightarrow \quad 8 - a>0 \quad \Rightarrow \quad a<8.
\]
Therefore, the least value of $a$ such that $f(x)$ is increasing on $[2, 4]$ is \( a = 8 \).
